Can someone find me an example of a continuous function which maps a closed set which is not open to an open set?
Let $S^1$ the unit circle and $f:\mathbb R\to S^1$, defined by $$ f(t)=(\cos t,\sin t). $$ Then $f[0,2\pi]=S^1$.
Here $S^1$ is open subset of itself and $[0,2\pi]$ is closed but not open.
Take $f:\Bbb R\rightarrow \Bbb R$, defined by $ f(x)=x\sin x $.
$f$ maps the set $[0,\infty)$ onto $\Bbb R$.
Let $A=[0,\infty)$ which is a closed but not open subset of $(-1/2,\infty).$ Map $A$ onto the open interval $(-1,1)$ by $f(x)=(x/(x+1))\sin(x).$ This wavers up and down as $x \to \infty$ to eventually pick up all of $(-1,1)$ in the image.
Let A be the graph of y = 1/x in R^2, and f be the projection map onto the first coordinate. Then A is closed and its image is open.